How many subsets of the set of divisors of 72 contain only composite numbers? For example, {8,9} and {4,8,12}$ are two such sets. Include the empty set in your count.

Respuesta :

Answer:

257

Step-by-step explanation:

The divisors of 72 which are composite number are

4,6,8,9,12,18,36,72 (composite numbers are those numbers which are formed by the multiplication of two whole numbers}

So the total number of divisors which are composite are = 8

So total number of subset [tex]=2^n=2^8=256[/tex]

In question it is said that also include empty set in count

So total number of subset =256+1=257

We know that the number of subsets of any given set is equal to $2^n,$ where $n$ is the number of elements in the set. First, then, we need to find the number of composite divisors. The prime factorization of $72$ is $72=2^3 \cdot 3^2,$ so there are $(3+1)(2+1)=12$ total divisors. (To see this, note that we can form a divisor of the form $2^a 3^b$ by freely choosing $a=0,1,2,3$ and $b=0,1,2$). Of these, $1$ is neither prime nor composite, and $2$ and $3$ are prime, for a total of $9$ composite divisors. Therefore there are $2^9=\boxed{512}$ subsets of the divisors of $72$ with only composite divisors.